Discussion Forum

A frame of the reference that is accelerated with respect to an inertial frame of reference is called a non- inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω is an example of a non-inertial frame of reference. The relationship between the force $\overrightarrow{F}_{rot}$ experienced by a particle of mass m moving on the rotating disc and the force $\overrightarrow{F}_{in}$ experienced by the particle in an inertial frame of reference is,

$\overrightarrow{F}_{rot}= \overrightarrow{F_{in}}+2m(\overrightarrow{v}_{rot}\times\overrightarrow{\omega})+ m(\overrightarrow{\omega}\times \overrightarrow{r})\times\overrightarrow{\omega}$

where $V_{rot}$ is the velocity of the particle in the rotating frame of reference and r is the position vector of the particle with respect to center of the disc

Now, consider a smooth slot along a diameter of a disc of radius R  rotating counter-clockwise with a constant angular speed ω about its vertical axis through its centre. We assign a coordinate system with the origin at the centre of the disc, the X-axis along the slot, the Y-axis perpendicular to the slot and Z-axis along the rotation axis $(\omega =\omega\hat{k})$. A small block of mass m is gently placed in the slot at $r=(\frac{R}{2})\hat{i}$ at t=0 and is constrained to move only along the slot

The distance r of the block  at time t is

Answer:

Explanation:

Force on block along slot=$m\omega^{2}r$

  $=ma=m(\frac{vdv}{dr})$

$\int_{0}^{v}vdv =\int_{R/2}^{r}\omega^{2} rdr$

$\Rightarrow\frac{v^{2}}{2}=\frac{\omega^{2}}{2}(r^{2}-\frac{R^{2}}{4})$

$v=\omega \sqrt{r^{2}-\frac{R^{2}}{4}}=\frac{\text{d}r}{\text{d}t}$

$\Rightarrow \int_{R/4}^{r} \frac{\text{d}r}{\sqrt{r^{2}-\frac{R^{2}}{4}}}=\int_{0}^{t}\omega dt $

$ln[\frac{r+\sqrt{r^{2}-\frac{R^{2}}{4}}}{\frac{R}{2}}$

- $ln[\frac{\frac{R}{2}and+\sqrt{\frac{R^{2}}{4}-\frac{R^{2}}{4}}}{\frac{R}{2}}]=\omega t$

$\Rightarrow r+\sqrt{r^{2}-\frac{R^{2}}{4}}=\frac{R}{2}e^{\omega t}$

$\Rightarrow r^{2}-\frac{R^{2}}{4}=\frac{R^{2}}{4}e^{2\omega t}+r^{2}-2r\frac{R}{2}e^{\omega t}$

$\Rightarrow r=\frac{\frac{R^{2}}{4}e^{2\omega t}+\frac{R^{2}}{4}}{Re^{\omega t}}=\frac{R}{4}(e^{\omega t}+e^{-\omega t})$